Question

Prove cauchy's triangle theorem

$\large\rm { \displaystyle\int_{ \Delta} f(z) dz = 0}$​

Answer 1

Suppose the contrary that,

$\large\rm{ I_{0} := \displaystyle\int_{ \Delta} f(z) dz \neq 0}$

Let A,B,C denotes the vertices of ∆ in the order following its orientation and let L,M and N denote the midpoints of the sides AB,BC, and CA, respectively. Form the four smaller oriented congruent triangles, $\large\rm{ \delta_{01} }$ = ALN, $\large\rm{ \delta_{02}}$ = LBM , $\large\rm{ \delta_{03}}$ = NMC and $\large\rm{ \delta_{04}}$ = MNL.

$\large\rm{ \Delta = ABC = \displaystyle\sum_{k=1}^{4} \delta_{0k} , }$

$\large\rm{ I_{0} = \displaystyle\sum_{k=1}^{4} \displaystyle\int_{ \delta 0k} f(z) dz}$

Let I1 denote the integral on the right-hand side with largest modulus and let ∆1 denote the corresponding triangle (one of the δ_0k). If two or more integrals share the same largest modulus, choose the one with lowest k. Then, according to the triangle inequality for complex numbers,

$\large\rm{| I_{0} | \leq \displaystyle\sum_{k=1}^{4} \Bigg | \displaystyle\int_{\delta \_ 0k} f(z) dz \Bigg | \leq 4 | I_{1} | }$

Next, the triangle ∆1 can be subdivided in the same manner into four congruent triangles δ1k, k = 1, 2, 3, 4, and one of those triangles ∆2, around which the integral of f(z) is I2, can be selected so that

$\large\rm{ |I_{1}| \geq 4 | I_{2} | }$

If the perimeter of ∆ is L0, the perimeter of ∆n is Ln = 2^{-n} L0. Let En denote the closed set consisting of the triangle ∆n and its interior. The sequence of closed sets, E0, E1, E2, ..., satisfies

$\large\rm{ E_{0} \supset E_{1} \supset E_{2} \supset \dots \supset E_{n} \dots }$

with $\large\rm{ | \eta (z) | < \epsilon |}$For sufficiently large n, the triangle ∆n and all later triangles in the sequence are contained within the open disc |z − z0| < δ. Since z0 is inside or on ∆n, this triangle will always be inside the disc if Ln < 2δ. Now

$\large\rm{ I_{n} = \displaystyle\int_{ \Delta n} \eta(z)(z- z_{0}) dz}$

where we used Lemma 2.11 to set the integrals of 1 and z to zero. In the last integrand, we have the bounds, |η(z)| < ǫ and |z −z0| < ½Ln, for $\large\rm{ z \in \Delta n}$. Hence, according to the ML formula,

$\large\rm{ | I_{0} \geq 2^{2n} | I_{n} | \geq \frac{1}{2} \epsilon ( L_{0} )^{2} }$

Because epsilon is arbitrary, the right-hand side is arbitrarily small. This contradicts our initial assumption that I0= 0. Hence I0 = 0. In other words,

$\large\rm{ \displaystyle\int_{R} f(z) dz = 0}$